Question

The tangent to the curve $y = e^x$ drawn at the point $(c, e^c)$ intersects the line joining the points $(c - 1, e^{c-1}) $ and $(c + 1, e^{c + 1})$

• A. on the left of x = c
• B. on the right of x = c
• C. at no point
• D. at all points

Answer 1

Answer: (A)

The equation of tangent to the curve $y = e^x$ at $(c ,e^c)$ is $y - e^c = e^c (x-c) $ .......(1)
and equation of line joining $\left(c-1, e^{c-1}\right)$ and $ \left(c+1, e^{c+1}\right) $
$ y-e^{c-1} = \frac{e^{c+1}-e^{c-1}}{\left(c+1\right)-\left(c-1\right)} \left[x-\left(c-1\right)\right] $
$\Rightarrow y -e^{c-1} = \frac{e^{c}\left(e-e^{-1}\right)}{2} \left[x-c+1\right] $ ......(2)
Subtracting equation (1) from (2), we get
$e^{c} - e^{c-1} = e^{c} \left(x-c\right) \left[\frac{e-e^{-1}-2}{2}\right]+e^{c} \left(\frac{e-e^{-1}}{2}\right)$
$ \Rightarrow x -c = \frac{\frac{\left[1-e^{-1} - \left(\frac{e-e^{-1}}{2}\right)\right]}{e-e^{-1}-2}}{2} = \frac{2-e-e^{-1}}{e-e^{-1}-2}$
$ = \frac{e+e^{-1}-2}{2-\left(e-e^{-1}\right)} $
$ = \frac{\frac{e+e^{-1}}{2} -1}{1- \frac{e-e^{-1}}{2}} = \frac{+ve}{-ve} = - ve $
$ \Rightarrow x -c < 0 \Rightarrow x < c $
$\therefore $ The two lines meet on the left of line x = c.